# Will a 4 Ah 12 V battery be fine for a 2 A 12 V diaphragm pump + a Raspberry Pi?

I've got a 4 Ah 12 V lead acid battery which currently runs a Raspberry PI. I'd like to connect a diaphragm pump to this battery as well. The pump has a max. rating of 2 A.

Assuming a fully charged battery : when the pump starts working, will this (a 4 Ah 12 V lead acid battery) be able to provide a steady voltage, or will the additional load (caused by the pump) cause undesired results? The pump will typically run for 10 - 15 minutes a day.

The Raspberry PI is apparantly notorious for being picky about the power supply.

If a 12 V 4 Ah battery is not enough; will it help if I replace the current battery with a larger one?

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Extra information:

The context of this question is that it is asked by someone who knows software development, but not electronics.

I used a RPi because that's what I had lying around and it runs Linux and Java. Easy to get started with.

The battery gets charged up every day via solar cells.

• What is "enough"? Is running for one day on one charge "enough" or do you need this to be able to run for a month on one charge? The RPi isn't picky about its power supply, my guess is such issues are only experienced by people who don't understand what an RPi needs. Just use an LM2596 based buck converter module to make 5 V from 12 V efficiently. BTW an RPi seems overkill for this function, an Arduino will be easier to configure for low power consumption. If I said you need a 12 V 200 Ah battery would you just take my word for it or would you ask why? – Bimpelrekkie May 13 at 8:09
• @Bimpelrekkie the context of this question is that it is asked by someone who knows software development, but not electronics. It should probably have written. So, I used a RPi because that's what I had lying around and it runs Linux and Java. Easy to get started with. The battery gets charged up every day via solar cells. I take your comment as "this should not be a problem" :-) – sbrattla May 13 at 8:13
• 1. Is this system located in Kristiansand (or geographically similar location? 2. How large is the PV panel. 3. It's quite likely that a much lower powered low cost Arduino (clones under $5 each) would do the job easily and it's likely that programming one would be easy enough for you. – Russell McMahon May 14 at 6:59 • ... 4. If you are repeatedly discharging the battery then a low depth of discharge is desirable for long life. Full discharge daily for the battery you cited may destroy it in 100-200 discharge cycles (maybe less). Aiming at only about 20% discharge is wise or buying a "deep discharge" rated battery - which sti;ll; benefit from far less than 100% DoD cycles. – Russell McMahon May 14 at 7:03 ## 2 Answers will this [battery] be able to provide a steady voltage It has a "Cold Cranking Current" rating of 60A, so a pump drawing e.g. 5A starting current will be absolutely no problem. Since the Pi needs a 5V supply, you need some kind of voltage step-down from the 12V+ battery anyway. A common step-down converter may operate on input down to about 2V above output voltage, so to about 7V for the Pi's 5V. That means that, as long as the battery's voltage does not drop below about 7V during pump start, you will be fine. And a charged 60A CCC battery in healthy condition should never drop to 7V from a 5A or even 10A current draw. As to the run times a quick estimation: Pump: 0.25h/day * 2A = 0.5Ah/day ~ the battery's 4Ah are enough for about 4Ah/0.5Ah/day = 8 days of pump operation. The Raspberry Pi, if running 24/7, will however consume much more energy: Optimistically assuming maybe 3W on average, (realistically probably more like 4-6W) that'd be 3W*24h/day = 72Wh/day required for the Pi alone. The battery holds (somewhat less usable energy than) 12V*4Ah = 48Wh So the Pi alone will drain the battery in (much) less than 2/3 of a day (16h). As others have stated, the Pi is probably way over the top for what you need. A small microcontroller can do many measurement, control and timing tasks as well or even better than a Pi (integrated ADC, less jitter in real-time applications, power save modes with wake-up 'cost' of a few micro- to milli-seconds,...). An AVR ATmega328 (the one also used in the Arduinos) draws about 30mA (at 5V, i.e. 0.15W) when busy. That's already down by a factor of 20-40x from an idle Pi, and, when not busy, can sleep at a fraction of that in deep power save modes. If you need WiFi you can go for the ESP8266 or the ESP32. They will draw considerable current (couple of hundred mA's) in bursts when "busy" sending/receiving data over the air, but can also use power save modes of a few mA. • Thanks a lot! Need to get a bigger battery for the rpi then, but great answer. Thanks! – sbrattla May 13 at 13:50 • @sbrattla If the rpi is overpowered for your application you could consider replacing it with a lower power microcontroller. You might be able to drop the power requirements by a few orders of magnitude. – Matt May 13 at 15:32 • Pi's get used because they are available. They are hardly ever a good choice for an actual application, the design is engineered to be plugged into the mains and to drive an HDMI TV; it's not designed as an embedded system, as it lacks both power management and boot media more robust than an SD card. – Chris Stratton May 13 at 16:02 • +1 Excellent answer. Addresses various points liable to be glossed over in typical answers. | Maybe some extra on "charged daily" requirements would be useful - but as that's in his comments it's not part of the question proper. – Russell McMahon May 14 at 4:24 • The OP stated (originally only in comments) that the battery will be charged daily "from solar cells". If he's in Southern Norway (as his profile suggests) then he'll need somewhat more than 500W of PV panels to power the system on an average December day!!! – Russell McMahon May 14 at 4:41 This is intended to complement JimmyB's excellent answer. Solar insolation in sometimes-sunny Kristiansand in the far South of Norway is shown in the table below taken from this Gaisma page - Kristiansand, Norway The equivalent full mean sunshine hours per day by moth are shown in the kWh/m^2/day entry. This shows that on a typical day from May to August you can expect ~= 4 to 5 SSH/day, BUT in mid Winter this falls to well under 1 SSH/day - 0.27 kWh/day/m^2 in December - if you keep the panels snow free. Jimmy B calculates that you need 6 + 48 = 54 Wh/day for pump + Pi if the Pi runs 24/7 at fullish power. A rule of thumb for starting solar power estimates is that you get about 50% of the energy out of a battery that the panel would provide in full available sun, properly pointed, clean and snow free. In this case if you are in Kristiansand in Norway (as I'm assuming) then in Summer you will ON AVERAGE get 5 SSH (= Sunshine hours = kWh/m^2/day) from your panels. BUT in December an average of 0.27 SSH/day. This means that to get 54 Wh/day you will need: • In summer 54 Wh x 2 loss factor / 5 hours ~= 22 W PV panel • In winter 54 Wh x 2 x 1/0.27 = 400 Watts of PV panel. The winter result is for an average December day with the panel snow free and pointed at about winter optimum angle. It assumes that your panel operates well in diffuse sun conditions (this varies). If you want to have enough energy for a run of even lower sun days you'd need to add extra battery capacity and extra PV capacity. That's a rather challenging environment for solar powering. Wind Turbine? A wind turbine (or snow mass turbine ? :-) ) may be better in deep winter. A wind turbine may indeed be a serious possibility. 48 Wh/day = 2W mean = trivial for a WT IF you can keep it snow free and intact in worst case winds. A mean wind speed of 3 m/s (11 km/h) will provide about 3 watts per m^2 of swept area at easily achievable efficiency figure of 20%. A 2m diameter 20% efficient WT will deliver 2 3 Watts in a 2 m/S (7 km/h) wind. The blades can be relatively light weight as long as the design can tolerate high winds when necessary. ____________________________________ It's quite likely that a much lower powered low cost Arduino (clones under$5 each) would do the job easily and it's likely that programming one would be easy enough for you.

If you are repeatedly discharging the battery then a low depth of discharge is desirable for long life. Full discharge daily for the battery you cited may destroy it in 100-200 discharge cycles (maybe less). Aiming at only about 20% discharge is wise or buying a "deep discharge" rated battery - which still benefit from far less than 100% DoD cycles.

• Good additional information and suggestions! - Just the 48 Wh/day for the Pi is too low; I actually guestimated 72 Wh/day in my answer :) - And I think this was also overly optimistic; 100Wh/day is probably about as low as one will get, depending on the model of Raspberry used, of course. – JimmyB May 14 at 14:43
• Thanks @Russell McMahon! Yes, I'm in Kristiansand and you're right about solar panels during the winter months. However, I'm using the rpi to run a hobbyist automatic watering system in my greenhouse (summer months). The power requirements you outline here do put things in perspective though! I did start using an rpi because its available and very easy to get started with. However, as you mention, i'm sure i'd be able get get this up and running on an arduino as well. I will have to try that this winter. The solar panels i've got is 18v 0.5A...😄 – sbrattla May 14 at 20:39
• @JimmyB i"m using the most power hungry rpi of them all (3b+?). I belive this does not improve he prospect of surviving on the current battery and solar panels... – sbrattla May 14 at 20:44
• @sbrattla 18v x 0.5A = 9W in peak sun. Orientation to sun for a fixed panel varies across day. Switching converter PV to RPi (18V to 5V) efficiency, and another (probably) to charge the battery and battery to Pi converter (maybe use the PV to Pi one when the light level drops) all lose energy. Battery charge and discharge lose energy. This is why I suggest a 50% efficincy PV out to load in overall until you have better figures. So your 9W panel becomes effectively 4.5W for load driving. To provide 54 Wh/day you need 54/4.5W ~= 12 hours sun - "which we have not got" even in mid summer. ... – Russell McMahon May 15 at 8:56
• ... To work on the 5+ hours of effective full sun May-July you'd need 54 Wh/(9W ) = 6h full sun so 120% efficiency from panel to load !!! :-) :-(. ie you need a bigger panel - even in mid summer! Eg 54Wh/5hsun = 11 Watts effective so a 20W panel would probably work from May to July and maybe to August. Wow! || A wind turbine is a serious suggestion. How is your wind situation? ... – Russell McMahon May 15 at 9:04